A partial ordering on a set S is a relation ≤ on S × S, satisfying the following.

(Reflexive) For any , we have .

(Transitive) For any , if and , then .

(Anti-symmetric) For any , if and , then .

A total ordering is a partial ordering such that for any , either or .

A partially ordered set (or just poset) is a set together with an assigned partial ordering. Similarly, we have totally ordered sets.

Given a partially ordered set , for , we write:

if and ;

if ;

if .

Examples

The set of real numbers (or rational numbers, or integers) forms a totally ordered set under the usual arithmetic ordering.

If X is a set, let P(X) be the collection of all subsets of X. This forms a partially ordered set under inclusion. If X has more than one element, P(X) is not totally ordered.

Any subset T of a partially ordered set S gives a partially ordered set. If S is totally ordered, so is T.

Lemma 1 (Duality).

If is a poset, then so is .

Proof.

Easy exercise. ♦

Duality allows us to cut our work by half in a lot of cases.

Bounds

Let be a poset.

Definition.

The maximum of S is an such that for any we have .

The minimum of S is an such that for any we have .

A maximal element of S is an such that for any , if then .

A minimal element of S is an such that for any , if then .

The naming is a little confusing, but one must differentiate between the maximum of a set and the maximal elements of a set. In the poset below, S has three maximal elements but no maximum.

For example, let X = {a, b, c} and let S be the following subsets of P(X) under inclusion.

The following properties are obvious.

Lemma 2.

The maximum (resp. minimum) of a poset is unique if it exists.

If the maximum (resp. minimum) of a poset exists, it is the unique maximal (resp. minimal) element.

Proof

Easy exercise. ♦

Exercise

Suppose S has a unique maximal element . Must be the maximum of S?

Finally, for a subset T of an ordered set S, we define the following.

Definition.

An upper bound (resp. lower bound) of T in S is an satisfying: for all , we have (resp. ).

Clearly, upper and lower bounds are not unique in general. E.g. for under the arithmetic ordering, the subset T of even integers has no upper or lower bound. The subset T’ of squares has lower bounds 0, -1, -2, … but no upper bound.

Noetherian Sets

This will be used a few times in our study of commutative algebra.

Definition.

A poset S is said to be noetherian if every non-empty subset T of S has a minimal element (in T).

It is easy to see that every finite poset S is noetherian.

Let be any non-empty subset. Pick . If is a minimal element of T we are done; otherwise there exists , . Again if is a minimal element of T we are done; otherwise we repeat. The process terminates since T is finite because we cannot have in T. Thus T has a minimal element.

Examples

The set of positive integers under ≤ is noetherian.

The set is noetherian, where if and .

The set is noetherian, where we take to mean .

Note that examples 2 and 3 are not totally ordered. The key property of noetherian sets is the following.

Theorem (Noetherian Induction).

Let T be a subset of a noetherian poset S satisfying the following.

If is such that , then .

Then T = S.

Note

To paraphrase the condition in words: if T contains all elements of S smaller than x, then it must contain x itself.

Proof

If , is non-empty so it has a minimal element x. By minimality, any with cannot lie in U so . But by the given condition this means , which is a contradiction. ♦

Here is an equivalent way of expressing the noetherian property.

Proposition 1.

A poset S is noetherian if and only if the following hold.

For any sequence of elements in S, there is an n such that .

Proof

(⇒) Suppose S is noetherian and are elements of S. The set thus has a minimal element . Since we have , equality must hold by minimality.

(⇐) Suppose S is not noetherian; let be a non-empty subset with no minimal element. Pick ; it is not minimal, so we can find with . Again since is not minimal we can find with . Repeat to obtain an infinitely decreasing sequence. ♦

Zorn’s Lemma

Finally we have the following critical result.

Theorem (Zorn’s Lemma).

Let S be a poset. A chain in S is a subset which is totally ordered. Suppose S is a poset such that every chain has an upper bound in S.

Then S has a maximal element.

A typical application of Zorn’s lemma is the following.

Proposition 2.

Every vector space V over a field k has a basis.

Note

This looks like an intuitively obvious result, but try finding a basis for the ℝ-space of all real functions . Using Zorn’s lemma, one can show that a basis exists but describing it does not seem possible. Generally, results that require Zorn’s lemma are of this nature: they claim existence of certain objects or constructions without exhibiting them explicitly. Some of these are rather unnerving, like the Banach-Tarski paradox.

Proof

Let be the set of all linearly independent subsets of V. We define a partial order on by inclusion, i.e. for , if and only if . Next, we claim that every chain has an upper bound.

Let ; we need to show that D is a linearly independent subset of V, so that is an upper bound of . Suppose

For each , we have for some . But , being a chain, is totally ordered so without loss of generality, we assume for all . This means ; since is linearly independent, we have .

Now apply Zorn’s lemma and we see that has a maximal element C. We claim that a maximal linearly independent subset of V must span V. If not, we can find outside the span of C. This means is linearly independent, thus contradicting the maximality of C. ♦

Zorn’s lemma holds vacuously when S is empty. However, in many applications of the lemma, constructing an upper bound for a chain requires T to be non-empty. In such instances, we will mentally replace Zorn’s lemma with the following equivalent version.

If S is a non-empty poset such that every non-empty chain in S has an upper bound in S, then S has a maximal element..